A solution of acetic acid is labeled 15%(w/v). How much of water must be added to 30.0 mL of this solution to reduce the concentration to 5.0%(w/v)? (Assume volumes are additive.)
A) 30.0 mL
B) 60.0 mL
C) 90.0 mL
D) 120.0 mL

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of acetic acid present in the initial solution:

15%(w/v) = 15 g of acetic acid in 100 mL of solution
Thus, in 30.0 mL of solution, there is:
(15 g/100 mL) x 30.0 mL = 4.5 g of acetic acid

Next, we need to calculate the volume of the diluted solution after water has been added:

Let V be the final volume of the diluted solution.
We know that the final concentration is 5%(w/v), so:
5%(w/v) = 5 g of acetic acid in 100 mL of solution

Since we are diluting the original solution by adding water, we can write the equation:
4.5 g of acetic acid / 30.0 mL = 5 g of acetic acid / V

Solving for V, we get:
V = (5 g of acetic acid x 30.0 mL) / 4.5 g of acetic acid
V = 33.3 mL

Therefore, we need to add:
V - 30.0 mL = 33.3 mL - 30.0 mL = 3.3 mL of water

Or, in other words, we need to add approximately 11% (3.3 mL / 30.0 mL x 100%) of the initial volume of solution in water. The total volume of the diluted solution will be:
30.0 mL + 3.3 mL = 33.3 mL

The closest answer choice is B) 60.0 mL, which is twice the volume of the initial solution. This means that we are adding the same amount of water as the amount of the initial solution, so the final volume will be:
30.0 mL + 30.0 mL = 60.0 mL

When we add 30.0 mL of water to the initial solution, the volume of the diluted solution becomes:
30.0 mL + 30.0 mL = 60.0 mL

We can calculate the final concentration by using the equation:
% (w/v) = (mass of solute / volume of solution) x 100

The mass of acetic acid in the diluted solution is still 4.5 g, but the volume is now 60.0 mL, so:
% (w/v) = (4.5 g / 60.0 mL) x 100
% (w/v) = 7.5%

This is not exactly 5%, but the closest answer choice is still B) 60.0 mL.

A Solution of Acetic Acid Is Labeled 15%(w/v)